Question

Let \(\overset{̄}{a},\overset{̄}{b},\overset{̄}{c}\) be three vectors of equal magnitude such that the angle between \(\overset{̄}{a}\) and \(\overset{̄}{b}\) is \(α\), \(\overset{̄}{b}\) and \(\overset{̄}{c}\) is \(β\), \(\overset{̄}{c}\) and \(\overset{̄}{a}\) is \(γ\).
Then the minimum value of \(cosα+cosβ+cosγ\) is ...

• A. \(\frac{1}{2}\)
• B. \(-\frac{1}{2}\)
• C. \(\frac{3}{2}\)
• D. \(-\frac{3}{2}\)

Hint:

A square of a vector sum is never negative.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Let each vector have magnitude \(m\). Then \(|\bar a + \bar b + \bar c|^2 \geq 0\).

Step 2: Key Formula or Approach:
\(|\bar a + \bar b + \bar c|^2 = 3m^2 + 2m^2(\cos\alpha + \cos\beta + \cos\gamma)\).

Step 3: Detailed Explanation:
Since this is at least \(0\): \(3 + 2(\cos\alpha + \cos\beta + \cos\gamma) \geq 0\).
So \(\cos\alpha + \cos\beta + \cos\gamma \geq -\frac32\).
Equality holds when \(\bar a + \bar b + \bar c = \bar 0\), for example three coplanar vectors at \(120^{\circ}\) to each other, where each cosine is \(-\frac12\).
\[ \text{Minimum} = -\frac32 \]

Final Answer:
The minimum value is \(-\frac{3}{2}\), option (D). \[ \boxed{-\frac{3}{2}} \]